A) \[\frac{\Delta (b,c,d)}{\Delta (a,b,c)}\]
B) \[\frac{-\Delta (b,c,d)}{\Delta (a,b,c)}\]
C) \[\frac{\Delta (a,c,d)}{\Delta (a,b,c)}\]
D) \[-\frac{\Delta (a,b,d)}{\Delta (a,b,c)}\]
Correct Answer: A
Solution :
[a] From the given system of equations, \[x=\frac{{{D}_{1}}}{D},\] \[y=\frac{{{D}_{2}}}{D},\] \[z=\frac{{{D}_{3}}}{D}\] where, \[D=\Delta (a,b,c);\] \[{{D}_{1}}=\Delta (d,b,c)\] \[{{D}_{2}}=\Delta (a,d,c);\] \[{{D}_{1}}=\Delta (a,b,d)\] Now, \[x=\frac{\Delta (d,b,c)}{\Delta (a,b,c)}\] where, \[\Delta (d,b,c)=\left| \begin{matrix} -{{d}_{1}} & {{b}_{1}} & {{c}_{1}} \\ -{{d}_{2}} & {{b}_{2}} & {{c}_{2}} \\ -{{d}_{3}} & {{b}_{3}} & {{c}_{3}} \\ \end{matrix} \right|\] \[=-\left| \begin{matrix} {{b}_{1}} & -{{d}_{1}} & {{c}_{1}} \\ {{b}_{2}} & -{{d}_{2}} & {{c}_{2}} \\ {{b}_{3}} & -{{d}_{3}} & {{c}_{3}} \\ \end{matrix} \right|=+\left| \begin{matrix} {{b}_{1}} & {{c}_{1}} & -{{d}_{1}} \\ {{b}_{2}} & {{c}_{2}} & -{{d}_{2}} \\ {{b}_{3}} & {{c}_{3}} & -{{d}_{3}} \\ \end{matrix} \right|\] \[=\Delta (b,c,d)\] Hence, \[x=\frac{\Delta (b,c,d)}{\Delta (a,b,c)}\]
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